Question

12. A boy kicks a rock off a cliff with a speed of 15.6 m/s at an angle of 52.3° above the horizontal. The rock hits the ground 5.82 s after it was kicked.

(12b) What is the speed of the rock right before it hits the ground?

(12c) What is the maximum height of the rock in the air, measured from the top of the cliff?

Please answer both b) and c).

Answer 1

Solution) initial velocity , Vo = 15.6 m/s

Angle (theeta) = 52.3°

Time t = 5.82 s

(b) speed of the rock right before it hits the ground , Vf = ?

Velocity along horizontal direction , Vh = (Vo)cos(theeta)

Vh = 15.6cos(52.3) = 9.54 m/s

Velocity along vertical direction , Vv = (Vo)sin(theeta)

Vv = 15.6sin(52.3) = 12.34 m/s

Along vertical direction

Vfy = ((Vo)sin(theeta))(t) - gt

Vfy = (15.6sin(52.3))(5.82) - 9.8×5.82 = 14.8 m/s

Along horizontal direction

Vfx = ((Vo)cos(theeta))(t) - at

In horizontal direction acceleration a = 0

Vfx = (15.6cos(52.3))(5.82)

Vfx = 55.52 m/s

Vf^2 = Vfx^2 + Vfy^2

Vf^2 = (55.52^2) + (14.8^2)

Vf^2 = 3301.51

Vf = (3301.51)^(1/2)

Vf = 57.45 m/s

(c) Maximum height , H = ?

We have ,

Vy^2 = (Voy)^2 - 2gd

At maximum height final velocity Vy = 0

0 = (Vosin(theeta))^2 - 2gd

d = (Vosin(theeta))^2/(2g)

d = ((15.6sin(52.3))^2)/(2×9.8)

d = 7.77 m

Maximum height measured from top of the cliff , H = d = 7.77 m

H = 7.77 m